Z xf X x dx and the variance is as before Var X E X 2 E X 2 where E X 2 Z x 2 f

Z xf x x dx and the variance is as before var x e x 2

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) = Z -∞ xf X ( x ) d x . and the variance is (as before) Var ( X ) = E ( X 2 ) - E ( X ) 2 , where E ( X 2 ) = Z -∞ x 2 f X ( x ) d x . By Arthur Mpazi Yambayamba Probability Distributions for Continuous Random Variables September 29, 2015 10 / 12
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The Probability Density Function The support or range of a continuous random variable is the smallest interval containing all values of x where f X ( x ) > 0 . Suppose that a random variable X has a pdf given by: f X ( x ) = ( 2 x if 0 x 1 0 otherwise The support of X is the interval [0,1]. We check the integral: Z -∞ f X ( x ) d x = Z 1 0 2 x d x = [ x 2 ] x =1 x =0 = 1 We have E ( X ) = Z -∞ xf X ( x ) d x = Z 1 0 2 x d x = Z 1 0 x * 2 x d x = Z 1 0 2 x 2 d x = 2 x 3 3 x =1 x =0 = 2 3 By Arthur Mpazi Yambayamba Probability Distributions for Continuous Random Variables September 29, 2015 11 / 12
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The Probability Density Function E ( X 2 ) = Z -∞ x 2 f X ( x ) d x = Z 1 0 x 2 * 2 x d x = Z 1 0 2 x 3 d x = 2 x 4 4 x =1 x =0 = 1 2 Var ( X ) = E ( X 2 ) - E ( X ) 2 = 1 2 - 2 3 2 = 1 18 The pdf of a random variable X is given by: f X ( x ) = ( cx if 0 x 4 0 otherwise (1) Find the value of c . (2) Hence, find the mean and variance of X . By Arthur Mpazi Yambayamba Probability Distributions for Continuous Random Variables September 29, 2015 12 / 12
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  • Fall '18
  • F. TAILOKA
  • Probability distribution, Probability theory, probability density function, continuous random variable Y, Arthur Mpazi Yambayamba

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